Need help figuring out what these graphs would look like.

Question

(a) is flowing at a constant rate (constant volume per unit time) into a cylindrical container standing vertically. Sketch a graph showing the depth of water against time. 

(b)Water is flowing at a constant rate into a cone shaped container standing on its point. Sketch a graph showing the depth of the water against time. 

Answer 1

A) The volume of the cylinder is

 

V=pi*r²h 

 

Take the derivative of this equation with respect to time in which is:

 

dV/dt = pi*r²*dh/dt  Note: The only thing that changes with time in filling the cylinder is the height of the cylinder.

 

This implies dh/dt=1/(pi*r²)*dV/dt.

 

Take the integral of both sides with respect to time in which is

 

∫dh/dt*dt = ∫1/(pi*r²)*dV/dt*dt This implies,

 

h(t) - h₀ = 1/(pi*r²)*dV/dt *(t - t₀) Note: Since the flow rate is constant, the right side of the equation gets multiplied by t. Thus,

 

h(t) = 1/(pi*r²)*dV/dt *t + h₀ assuming t₀ = 0

 

Now, you got a linear equation in which the slope is 1/(pi*r2)*dV/dt and the initial height at time equal zero is h₀.

 

 b) The volume of the cone is:

 

V=1/3*pi*r²*h

 

Since r and h would vary with time, it would be good practice to rewrite the equation as a function of h since we are only interested in deriving an equation of height as a function of time. If you drew a graph representing the cone to determine the relation of r and h, you would get:

 

r(h)=(R/H)*h where R is the radius of the cone's base and H is its height. Substitute this in the equation of the volume of the cone.

 

V=1/3*pi*((R/H)*h)²*h = 1/3*pi*(R/H)²*h³

 

Now take the derivative of the equation with respect to time.

 

dV/dt = 1/3*pi*(R/H)²*3h²*dh/dt = pi*(R/H)²*h²*dh/dt

 

At this point, you got to be careful if you want to derive h(t). The multiply the equation by dt and divide it by pi*(R/H)², and with some algebraic manipulation the equation becomes:

 

h²dh = (H/R)²/pi *dV/dt*dt

 

Take the integral of both sides.

 

∫h²dh = ∫(H/R)²/pi *dV/dt*dt

 

This implies,

 

1/3*((h(t)³ - h₀³) = (H/R)²/pi *dV/dt*(t-t₀)

 

(h(t))³ = 3*(H/R)²/pi *dV/dt*(t-t0) + h₀³

 

Thus,

 

h(t) = (3*(H/R)²/pi *dV/dt*t+ h₀³)^1/3 Note: The whole expression on the right hand of the equation is taken to the 1/3 power.